A box, constructed from a rectangular metal sheet, is 21 cm by 16 cm by cutting equal squares of sides x from the corners of the sheet and then turning up the projected portions. The value of x so that volume of the box is maximum is

The dimensions of the box after cutting equal squares of side x on the corner will be 21 - 2x, 16 - 2x, and height x.
$\mathrm{V}=\mathrm{x}(21-2\mathrm{x})(16-2\mathrm{x})$
   $=\mathrm{x}\left(336-74\mathrm{x}+4{\mathrm{x}}^2\right)=4{\mathrm{x}}^3+336\mathrm{x}-74{\mathrm{x}}^2$
$\therefore \frac{\mathrm{dV}}{\mathrm{dx}}=12{\mathrm{x}}^2+336-148\mathrm{x}$
$\frac{\mathrm{dV}}{\mathrm{dx}}=0\text{ gives }\mathrm{x}=3\text{ for which }\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dx}}^2}\text{ is }-\mathrm{Ve}$ and, hence, maximum.